Trilogarithm $\operatorname{Li}_3(z)$ and the imaginary golden ratio $i\,\phi$

Question

I experimentally discovered the following conjectures: $$\Re\Big[1800\operatorname{Li}_3(i\,\phi)-24\operatorname{Li}_3\left(i\,\phi^5\right)\Big]\stackrel{\color{gray}?}=100\ln^3\phi-47\,\pi^2\ln\phi-150\,\zeta(3),\tag1$$ $$\Im\Big[720\operatorname{Li}_3(i\,\phi)-320\operatorname{Li}_3\left(i\,\phi^3\right)-48\operatorname{Li}_3\left(i\,\phi^5\right)\Big]\stackrel{\color{gray}?}=9\,\pi^3-780\,\pi\ln^2\phi,\tag2$$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio and $\operatorname{Li}_3(z)$ is the trilogarithm. They check numerically with at least $20000$ decimal digits. It appears that Maple and Mathematica know nothing about these identities.

Are these known identities? How can we prove them?

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Update: It also appears that $$\begin{align}&\Re\operatorname{Li}_3(i\,\phi)\stackrel{\color{gray}?}=\frac1{32}\operatorname{Li}_3\left(\phi^{-4}\right)+\frac3{16}\operatorname{Li}_3\left(\phi^{-2}\right)-\frac38\ln^3\phi-\frac14\zeta(3)\\ \,\\ &\Re\operatorname{Li}_3\left(i\,\phi^3\right)\stackrel{\color{gray}?}=\frac9{32}\operatorname{Li}_3\left(\phi^{-4}\right)+\frac12\operatorname{Li}_3\left(\phi^{-3}\right)+\frac38\operatorname{Li}_3\left(\phi^{-2}\right)-\frac38\operatorname{Li}_3\left(\phi^{-1}\right)-\frac{43}8\ln^3\phi-\frac{15}{32}\zeta(3)\\ \,\\ &\Re\operatorname{Li}_3\left(i\,\phi^5\right)\stackrel{\color{gray}?}=\frac{75}{32}\operatorname{Li}_3\left(\phi^{-4}\right)-\frac58\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{45}2\ln^3\phi-\frac34\zeta(3)\end{align}$$ These together with a known value for $\operatorname{Li}_3\left(\phi^{-2}\right)$ imply $(1)$.

Answer 1

(Too long for a comment.)

You may have missed the small even powers $k=2,4,6$ $$\begin{align} &32\,\Re\operatorname{Li}_3\left(i\,\phi^2\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-8}\right)-4\operatorname{Li}_3\left(\phi^{-4}\right)+20\operatorname{Li}_3\left(\phi^{-2}\right)-7\ln^3(\phi^2)-16\,\zeta(3)\\ \,\\ &32\,\Re\operatorname{Li}_3\left(i\,\phi^4\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-16}\right)-4\operatorname{Li}_3\left(\phi^{-8}\right)+40\operatorname{Li}_3\left(\phi^{-2}\right)-46\ln^3(\phi^2)-32\,\zeta(3)\\ \,\\ &32\,\Re\operatorname{Li}_3\left(i\,\phi^6\right)\stackrel{\color{blue}?}=18\operatorname{Li}_3\left(\phi^{-8}\right)-72\operatorname{Li}_3\left(\phi^{-4}\right)+64\operatorname{Li}_3\left(\phi^{-3}\right)+90\operatorname{Li}_3\left(\phi^{-2}\right)-48\operatorname{Li}_3\left(\phi^{-1}\right)-1196\ln^3\phi-35\,\zeta(3) \end{align}$$ More generally, for any real number $k$ it seems, $$32\,\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ though I have no proof yet.

Answer 2

I'll try to prove Tito Piezas III's (+1) neat conjecture : $$\tag{1}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ The classical identity $\;\displaystyle\frac {\operatorname{Li}_3(x^2)}4=\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\;$ (c.f. Lewin $1981$ "Polylogaritms and associated functions" p.$154$) allows to rewrite the two first terms at the right as : $$\tag{2}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ and gives further : $\;\dfrac {\operatorname{Li}_3\left(-\phi^{2k}\right)}4=\operatorname{Li}_3\left(i\,\phi^k\right)+\operatorname{Li}_3\left(-i\,\phi^k\right)=2\Re\operatorname{Li}_3\left(i\,\phi^k\right)\;$ that is : $$\tag{3}4\operatorname{Li}_3\left(-\phi^{2k}\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$

Another usual identity is $\;\displaystyle \operatorname{Li}_3(-x)-\operatorname{Li}_3(-1/x)=-\zeta(2)\log(x)-\frac{\ln^3(x)}6\;$ (same Lewin page)
applied to $\,x:=\phi^{2k}\,$ $(3)$ becomes : \begin{align} -8k\zeta(2)\ln(\phi)-\frac{4k^3\ln^3(\phi^{2})}6\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)\\ -8k\zeta(2)\ln(\phi)+\frac{5k\ln^3(\phi^{2})}6+8k\,\zeta(3)\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)\\ \tag{4}\operatorname{Li}_3\left(\phi^{-2}\right)=&\frac 45(\zeta(3)-\zeta(2)\ln(\phi))+\frac 23{\ln^3(\phi)}\\ \end{align} This last identity is proved in Lewin's $1991$ book (page $2$ with $\,\rho=\dfrac 1{\phi}\,$) and may be compared with alpha's evaluation knowing that $\,\operatorname{csch}^{-1}(2)=\ln(\phi)$.